# Eigenfunction Matching for a Finite Change in Depth

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## Introduction

The problem consists of a region of free water surface with depth $h$ except between $-L$ and $L$ where the depth is $d$. The problem with a semi-infinite change in depth is treated in Eigenfunction Matching for a Semi-Infinite Change in Depth

## Governing Equations

We begin with the Frequency Domain Problem for a submerged dock which occupies the region $x\gt 0$ (we assume $e^{-\mathrm{i}\omega t}$ time dependence). The depth of is constant $h$ for $x\lt 0$ and constant $d$ for $x\gt 0$. The $z$-direction points vertically upward with the water surface at $z=0$. The boundary value problem can therefore be expressed as

$\Delta\phi=0, \,\, -h\lt z\lt 0, \,\, x\lt -L,\,x\gt L$
$\Delta\phi=0, \,\, -d\lt z\lt 0, \,\, -L\lt x\lt L$
$\partial_z\phi=\alpha\phi, \,\, z=0,$
$\partial_x\phi=0, \,\, -d\lt z\lt -h,\,x=\pm L,$
$\partial_z\phi=0, \,\, z=-h,\, x\lt -L,\,x\gt L$
$\partial_z\phi=0, \,\, z=-d,\, -L\lt x\lt L$

We must also apply the Sommerfeld Radiation Condition as $|x|\rightarrow\infty$. This essentially implies that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave and a wave propagating away.

## Solution Method

We use separation of variables in the two regions, $x\lt 0$ and $x\gt 0$.

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

### Separation of variables for a free surface

We use separation of variables

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

The separation of variables equation for deriving free surface eigenfunctions is as follows:

$Z^{\prime\prime} + k^2 Z =0.$

subject to the boundary conditions

$Z^{\prime}(-h) = 0$

and

$Z^{\prime}(0) = \alpha Z(0)$

We can then use the boundary condition at $z=-h \,$ to write

$Z = \frac{\cos k(z+h)}{\cos kh}$

where we have chosen the value of the coefficent so we have unit value at $z=0$. The boundary condition at the free surface ($z=0 \,$) gives rise to:

$k\tan\left( kh\right) =-\alpha \,$

which is the Dispersion Relation for a Free Surface

The above equation is a transcendental equation. If we solve for all roots in the complex plane we find that the first root is a pair of imaginary roots. We denote the imaginary solutions of this equation by $k_{0}=\pm ik \,$ and the positive real solutions by $k_{m} \,$, $m\geq1$. The $k \,$ of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

$\cos ix = \cosh x, \quad \sin ix = i\sinh x,$

to arrive at the dispersion relation

$\alpha = k\tanh kh.$

We note that for a specified frequency $\omega \,$ the equation determines the wavenumber $k \,$.

Finally we define the function $Z(z) \,$ as

$\chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0$

as the vertical eigenfunction of the potential in the open water region. From Sturm-Liouville theory the vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but this has no advantages for a numerical implementation. It can be shown that

$\int\nolimits_{-h}^{0}\chi_{m}(z)\chi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn}$

where

$A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos ^{2}k_{n}h}\right).$

### Inner product between free surface and dock modes

$B_{mn} = \int\nolimits_{-h}^{0}\phi_{m}^{d}(z)\phi_{n}^{h}(z) \mathrm{d} z$

where

$B_{mn}= \int\nolimits_{-h}^{0} \frac{\cos(k_{m}^{h}(z+h))\cos(k_{n}^{d}(z+d))}{\cos(k_{m}h)\cos(k_{n}d)} \mathrm{d} z =\frac{k_{m}^{h}\sin(k_{m}^{d} h)\cos(k_{n}^{d}h)-k_{n}^{d}\cos(k_{m}^{h} h)\sin(k_{n}^{d}h)} {\cos(k_{m}h)\cos(k_{n}d)({k_{m}^{d}}^{2}-{k_{n}^{d}}^{2})}$

## Governing Equations

We begin with the Frequency Domain Problem for a submerged dock which occupies the region $x\gt 0$ (we assume $e^{i\omega t}$ time dependence). The depth of is constant $h$ for $x\lt -L$ and $x\gt L$ and constant $d$ for $-L\lt x\gt L$. The $z$-direction points vertically upward with the water surface at $z=0$. The boundary value problem can therefore be expressed as

$\Delta\phi=0, \,\, -h\lt z\lt 0, \,\, x\lt -L\,\textrm{or}\, x\gt L,$
$\Delta\phi=0, \,\, -d\lt z\lt 0, \,\, -L\lt x\lt L,$
$\partial_z\phi=\alpha\phi, \,\, z=0,$
$\partial_x\phi=0, \,\, -d\lt z\lt -h,\,x=\pm{L},$
$\partial_z\phi=0, \,\, z=-h,\, x\lt -L\,\textrm{or}\, x\gt L,$
$\partial_z\phi=0, \,\, z=-d,\, -L\lt x\lt L.$

We must also apply the Sommerfeld Radiation Condition as $|x|\rightarrow\infty$. This essentially implies that the only wave at infinity is propagating away and at negative infinity there is a unit incident wave and a wave propagating away.

## Solution Method

We use separation of variables in the two regions, $x\lt 0$ and $x\gt 0$.

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

### Separation of variables for a free surface

We use separation of variables

We express the potential as

$\phi(x,z) = X(x)Z(z)\,$

and then Laplace's equation becomes

$\frac{X^{\prime\prime}}{X} = - \frac{Z^{\prime\prime}}{Z} = k^2$

The separation of variables equation for deriving free surface eigenfunctions is as follows:

$Z^{\prime\prime} + k^2 Z =0.$

subject to the boundary conditions

$Z^{\prime}(-h) = 0$

and

$Z^{\prime}(0) = \alpha Z(0)$

We can then use the boundary condition at $z=-h \,$ to write

$Z = \frac{\cos k(z+h)}{\cos kh}$

where we have chosen the value of the coefficent so we have unit value at $z=0$. The boundary condition at the free surface ($z=0 \,$) gives rise to:

$k\tan\left( kh\right) =-\alpha \,$

which is the Dispersion Relation for a Free Surface

The above equation is a transcendental equation. If we solve for all roots in the complex plane we find that the first root is a pair of imaginary roots. We denote the imaginary solutions of this equation by $k_{0}=\pm ik \,$ and the positive real solutions by $k_{m} \,$, $m\geq1$. The $k \,$ of the imaginary solution is the wavenumber. We put the imaginary roots back into the equation above and use the hyperbolic relations

$\cos ix = \cosh x, \quad \sin ix = i\sinh x,$

to arrive at the dispersion relation

$\alpha = k\tanh kh.$

We note that for a specified frequency $\omega \,$ the equation determines the wavenumber $k \,$.

Finally we define the function $Z(z) \,$ as

$\chi_{m}\left( z\right) =\frac{\cos k_{m}(z+h)}{\cos k_{m}h},\quad m\geq0$

as the vertical eigenfunction of the potential in the open water region. From Sturm-Liouville theory the vertical eigenfunctions are orthogonal. They can be normalised to be orthonormal, but this has no advantages for a numerical implementation. It can be shown that

$\int\nolimits_{-h}^{0}\chi_{m}(z)\chi_{n}(z) \mathrm{d} z=A_{n}\delta_{mn}$

where

$A_{n}=\frac{1}{2}\left( \frac{\cos k_{n}h\sin k_{n}h+k_{n}h}{k_{n}\cos ^{2}k_{n}h}\right).$

## Solution using Symmetry

The finite dock problem is symmetric about the line $x=0$ and this allows us to solve the problem using symmetry. This method is numerically more efficient and requires only slight modification of the code for Eigenfunction Matching for a Semi-Infinite Dock, the developed theory here is very close to the semi-infinite solution. We decompose the solution into a symmetric and an anti-symmetric part as is described in Symmetry in Two Dimensions

### Symmetric solution

The symmetric potential can be expanded as

$\phi(x,z)=e^{-k_{0}^{h}(x+L)}\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m}^{s}e^{k_{m}^{h}(x+L)}\phi_{m}^{h}(z) , \;\;x\lt -L$

and

$\phi(x,z)=\sum_{m=0}^{\infty}b_{m}^{s} \frac{\cosh k_{m}^{d} x}{\cosh k_{m}^{d} L} \phi_{m}^{d}(z), \;\;-L\lt x\lt 0$

where $a_{m}^{s}$ and $b_{m}^{s}$ are the coefficients of the potential in the two regions.

For the first equation we multiply both sides by $\phi_{n}^{h}(z) \,$ and integrate from $-h$ and for the second equation we multiply both sides by $\phi_{n}^{d}(z) \,$ and integrate from $-d$. This gives us

$A_{0}\delta_{0n}+a_{n}^{s}A_{n}^{h} =\sum_{m=0}^{\infty}b_{m}^{s}B_{mn}$

and

$-k_{0}^{h}B_{n0} + \sum_{m=0}^{\infty} a_{m}^{s}k_{m}^{h} B_{nm} = -b_{n}^{s}k_{n}^{d}\tanh(k_{n}^{d}L) A_{n}^{d}$

(for full details of this derivation see Eigenfunction Matching for a Semi-Infinite Change in Depth)

### Anti-symmetric solution

The anti-symmetric potential can be expanded as

$\phi(x,z)=e^{-k_{m}^{h}(x+L)}\phi_{0}\left( z\right) + \sum_{m=0}^{\infty}a_{m}^{a}e^{k_{m}^{h}(x+L)}\phi_{m}(z) , \;\;x\lt -L$

and

$\phi(x,z)=\sum_{m=0}^{\infty}b_{m}^{a} \frac{\sinh k_{m}^{h} x}{-\sinh k_{m}^{h} L}\phi_{m}(z), \;\;-L\lt x\lt 0$

where $a_{m}^{a}$ and $b_{m}^{a}$ are the coefficients of the potential in the two regions. Note that the minus sign in the expression for the dock-covered region has been added so that each component is equal to one at $x=-L$.

For the first equation we multiply both sides by $\phi_{n}^{h}(z) \,$ and integrate from $-h$ and for the second equation we multiply both sides by $\phi_{n}^{d}(z) \,$ and integrate from $-d$. This gives us

$A_{0}^{h}\delta_{0n}+a_{n}^{a}A_{n}^{h} =\sum_{m=0}^{\infty}b_{m}^{a}B_{mn}$

and

$-k_{0}^{h}B_{n0} + \sum_{m=0}^{\infty} a_{m}^{a}k_{m}^{h} B_{nm} = -b_{n}^{a}k_{n}^{d}\coth(k_{n}^{d}L) A_{n}^{d}$

(for full details of this derivation see Eigenfunction Matching for a Semi-Infinite Change in Depth)

### Solution to the original problem

We can now reconstruct the potential for the finite dock from the two previous symmetric and anti-symmetric solution as explained in Symmetry in Two Dimensions. The amplitude in the left open-water region is simply obtained by the superposition principle

$a_{m} = \frac{1}{2}\left(a_{m}^{s}+a_{m}^{a}\right)$

$d_{m} = \frac{1}{2}\left(a_{m}^{s}-a_{m}^{a}\right)$

Note the formulae for $b_m$ and $c_m$ are more complicated but can be derived with some work.

## Matlab Code

A program to calculate the coefficients for the semi-infinite dock problems can be found here finite_change_in_depth.m

### Additional code

This program requires